Question: $ B = \left[\begin{array}{rrr}5 & 3 & 3 \\ 4 & 0 & 4\end{array}\right]$ $ E = \left[\begin{array}{rr}0 & 3 \\ -2 & 5 \\ 1 & -2\end{array}\right]$ What is $ B E$ ?
Solution: Because $ B$ has dimensions $(2\times3)$ and $ E$ has dimensions $(3\times2)$ , the answer matrix will have dimensions $(2\times2)$ $ B E = \left[\begin{array}{rrr}{5} & {3} & {3} \\ {4} & {0} & {4}\end{array}\right] \left[\begin{array}{rr}{0} & \color{#DF0030}{3} \\ {-2} & \color{#DF0030}{5} \\ {1} & \color{#DF0030}{-2}\end{array}\right] = \left[\begin{array}{rr}? & ? \\ ? & ?\end{array}\right] $ To find the element at any row $i$ , column $j$ of the answer matrix, multiply the elements in row $i$ of the first matrix, $ B$ , with the corresponding elements in column $j$ of the second matrix, $ E$ , and add the products together. So, to find the element at row 1, column 1 of the answer matrix, multiply the first element in ${\text{row }1}$ of $ B$ with the first element in ${\text{column }1}$ of $ E$ , then multiply the second element in ${\text{row }1}$ of $ B$ with the second element in ${\text{column }1}$ of $ E$ , and so on. Add the products together. $ \left[\begin{array}{rr}{5}\cdot{0}+{3}\cdot{-2}+{3}\cdot{1} & ? \\ ? & ?\end{array}\right] $ Likewise, to find the element at row 2, column 1 of the answer matrix, multiply the elements in ${\text{row }2}$ of $ B$ with the corresponding elements in ${\text{column }1}$ of $ E$ and add the products together. $ \left[\begin{array}{rr}{5}\cdot{0}+{3}\cdot{-2}+{3}\cdot{1} & ? \\ {4}\cdot{0}+{0}\cdot{-2}+{4}\cdot{1} & ?\end{array}\right] $ Likewise, to find the element at row 1, column 2 of the answer matrix, multiply the elements in ${\text{row }1}$ of $ B$ with the corresponding elements in $\color{#DF0030}{\text{column }2}$ of $ E$ and add the products together. $ \left[\begin{array}{rr}{5}\cdot{0}+{3}\cdot{-2}+{3}\cdot{1} & {5}\cdot\color{#DF0030}{3}+{3}\cdot\color{#DF0030}{5}+{3}\cdot\color{#DF0030}{-2} \\ {4}\cdot{0}+{0}\cdot{-2}+{4}\cdot{1} & ?\end{array}\right] $ Fill out the rest: $ \left[\begin{array}{rr}{5}\cdot{0}+{3}\cdot{-2}+{3}\cdot{1} & {5}\cdot\color{#DF0030}{3}+{3}\cdot\color{#DF0030}{5}+{3}\cdot\color{#DF0030}{-2} \\ {4}\cdot{0}+{0}\cdot{-2}+{4}\cdot{1} & {4}\cdot\color{#DF0030}{3}+{0}\cdot\color{#DF0030}{5}+{4}\cdot\color{#DF0030}{-2}\end{array}\right] $ After simplifying, we end up with: $ \left[\begin{array}{rr}-3 & 24 \\ 4 & 4\end{array}\right] $